Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

prove that the straight line x/a+y/b=1 touches the curve y=be^-x/a at that point where the curve cuts the y-axis also find the equation of normal at the same point

prove that the straight line x/a+y/b=1 touches the curve y=be^-x/a at that point where the curve cuts the y-axis also find the equation of normal at the same point 
 

Grade:12th pass

1 Answers

yathartha gupta
71 Points
3 years ago
Straight line touches the curve at y axis so there x will be 0Put the value in equationy=be^-0/ay=be^0Y=b.So the co-ordinates of the point will be(0,b)By diff. the curve equationY=be^-x/ady/dx=-b/ae^-x/a[dy/dx](0,b) = -b/ae^0 = -b/aSo the slope of normal will be a/bThe equation of normal at (0,b) will bey-y1=m(x-x1)y - b = a/b(x - 0)y - b = ax/by - ax/b = bby - ax =b^2..................Ans.

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free